Show that the points \( A(-5,6), B(3,0) \) and \( C(9,8) \) are the vertices of an isosceles triangle.

To determine whether the points \( A(-5, 6) \), \( B(3, 0) \), and \( C(9, 8) \) form an isosceles triangle, we'll calculate the distances between each pair of points. A triangle is isosceles if at least two of its sides are of equal length. ### Step 1: Calculate the Distance Between Each Pair of Points The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in the Cartesian plane is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Let's compute the distances: #### 1. Distance between \( A(-5, 6) \) and \( B(3, 0) \) — \( AB \) \[ \begin{align*} AB &= \sqrt{(3 - (-5))^2 + (0 - 6)^2} \\ &= \sqrt{(8)^2 + (-6)^2} \\ &= \sqrt{64 + 36} \\ &= \sqrt{100} \\ &= 10 \end{align*} \] #### 2. Distance between \( B(3, 0) \) and \( C(9, 8) \) — \( BC \) \[ \begin{align*} BC &= \sqrt{(9 - 3)^2 + (8 - 0)^2} \\ &= \sqrt{(6)^2 + (8)^2} \\ &= \sqrt{36 + 64} \\ &= \sqrt{100} \\ &= 10 \end{align*} \] #### 3. Distance between \( A(-5, 6) \) and \( C(9, 8) \) — \( AC \) \[ \begin{align*} AC &= \sqrt{(9 - (-5))^2 + (8 - 6)^2} \\ &= \sqrt{(14)^2 + (2)^2} \\ &= \sqrt{196 + 4} \\ &= \sqrt{200} \\ &= 10\sqrt{2} \approx 14.14 \end{align*} \] ### Step 2: Analyze the Results - \( AB = 10 \) - \( BC = 10 \) - \( AC = 10\sqrt{2} \) Since \( AB = BC \), two sides of the triangle are equal in length. The third side, \( AC \), is longer but does not affect the isosceles property, which requires only **two** sides to be equal. ### Conclusion The triangle formed by points \( A(-5, 6) \), \( B(3, 0) \), and \( C(9, 8) \) has two sides of equal length (\( AB = BC \)). Therefore, it is an **isosceles triangle**.